Linear Dependence in Inner Product Space Given an inner product space X, let x,y $\in$ X.
I need to show the following are equivalent:
$\|$x+y$\|$ = $\|$x$\|$ + $\|$y$\|$ $\iff$ $\|$y$\|$x=$\|$x$\|$y .
I know the first condition makes x and y linearly dependent but I haven't gone much further than the Cauchy-Schwarz inequality, triangle inequality and parallelogram law so I'm not sure how to get the "free" x and y on the right hand side.
 A: If $x$ is the zero vector, the equivalence holds. Suppose that $x$ is nonzero. Rewrite the left-hand side of the equivalence as
$$
\sqrt{(x+y,x+y)}=\sqrt{(x,x)}+\sqrt{(y,y)}
$$
and raise both parts to the square, thereby getting that
$$
(x,y)=|x|\,|y|, 
$$
or
\begin{equation*} \tag{1}
(x/|x|,y)=|y|.
\end{equation*} 
Consider then an orthonormal basis $x_1,\ldots,x_n$ such that $x_1=x/|x|$ and
let
$$
y=\alpha_1 x_1 + \ldots + \alpha_n x_n. 
$$ 
Then it follows from (1) that
$$
\alpha_1 = \sqrt{\alpha_1^2 + \ldots +\alpha_n^2},
$$
whence
$$
\alpha_1=|y|, \alpha_2=\ldots=\alpha_n=0.
$$
or
$$
y=|y| x/|x| \iff |x| y = |y| x. 
$$
The converse is trivial.
A: The stated condition is equivalent to the following conditions
$$
                   \|x+y\|^2 = (\|x\|+\|y\|)^2 \\
                    \|x\|^2+2\Re(x,y)+\|y\|^2=\|x\|^2+2\|x\|\|y\|+\|y\|^2 \\
               \Re(x,y) = \|x\|\|y\|.
$$
And the following are also equivalent
$$
                \|\|y\|x-\|x\|y\|^2 = 0 \\
              \|y\|^2\|x\|^2-2\|x\|\|y\|\Re(x,y)+\|x\|^2\|y\|^2 = 0 \\
                  2\|x\|\|y\|\{\|x\|\|y\|-\Re(x,y)\}=0.
$$
If $x=0$ or $y=0$ then $\|x\|y=\|y\|x$ and $\|x+y\|=\|x\|+\|y\|$. Otherwise, if neither is $0$, then
$$
      \|x+y\| = \|x\|+\|y\| \iff \|y\|x-\|x\|y = 0.
$$
A: There is no need to use bases. ||x+y|| = ||x||+||y|| implies x²+y²+2x•y = x²+y²+2||x||||y||, so 2x•y=2||x||||y||. Now write x=||x||*y/||y||+v, where v is the "error term". This becomes
2||x||y•y/||y|| + 2v•y = 2||x||||y||
So 2v•y = 0. Reversing that y=(x-v)*||y||/||x|| shows that v•x=0 as well. This means that v is in space of vectors normal to x and y. At the same time, v is in the subspace spanned by x and y. These are complementary, so v=0; done.
